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Ta có :
\(5^{199}< 5^{200}=5^{2\cdot100}=25^{100}\)
\(3^{300}=3^{3\cdot100}=27^{100}\)
Mà \(25^{100}< 27^{100}\Rightarrow5^{199}< 3^{300}\)
Vậy \(\dfrac{1}{3^{300}}>\dfrac{1}{5^{199}}\)
3³⁰⁰ = (3³)¹⁰⁰ = 27¹⁰⁰
5²⁰⁰ = (5²)¹⁰⁰ = 25¹⁰⁰
Do 27 > 5 nên 27¹⁰⁰ > 25¹⁰⁰
⇒ 3³⁰⁰ > 5²⁰⁰ (1)
Do 200 > 199 nên 5²⁰⁰ > 5¹⁹⁹ (2)
Từ (1) và (2) ⇒ 3³⁰⁰ > 5¹⁹⁹
⇒ 1/3³⁰⁰ < 1/5¹⁹⁹
a.\(\frac{-1}{5}+\frac{4}{2}.\frac{5}{2}-\sqrt{25}\)
= \(\frac{-1}{5}\)+5 - 5
\(\frac{-1}{5}\)
b. 23\(\frac{1}{4}\).\(\frac{7}{5}\)- 13\(\frac{1}{4}\):\(\frac{5}{7}\)
=\(\frac{651}{20}\)- \(\frac{371}{20}\)
= \(\frac{651-371}{20}\)
= 14
a) ( -2.5 ) . ( 7,5) .( -4 )
= [(-2,5).(-4)].(7,5)
= 10 . 7,5
= 75
b) \(1\frac{4}{23}+\frac{8}{21}-\frac{4}{23}+0,6+\frac{13}{21}\)
=\(1\frac{4}{23}-\frac{4}{23}+\frac{8}{21}+\frac{13}{21}-0,6\)
\(=1+1-0,6\)
\(=2-0,6\)
= 1,4
c) \(\frac{2}{7}.15\frac{1}{3}-\frac{2}{7}.20.\frac{1}{3}+4\frac{1}{3}\)
\(=\frac{2}{7}.5-\frac{1}{3}.\frac{40}{7}+4\frac{1}{3}\)
= \(=\frac{10}{7}-\frac{17}{7}\)
= -1
d) \(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)
\(=\frac{9}{4}.\left( \frac{-5}{3}\right)-\frac{5}{4}.\left(\frac{-5}{3}\right)\)
=\(\left(\frac{-5}{3}\right).\left(\frac{9}{4}-\frac{5}{4}\right)\)
\(=\frac{-5}{3}.1\)
\(=\frac{-5}{3}\)
\(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}.\dfrac{7}{5}\)
= \(\dfrac{7}{5}.\left(\dfrac{93}{4}-\dfrac{53}{4}\right)\)
= \(\dfrac{7}{5}.10\)
= \(14\)
1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=1+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}\)
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